# Difference between revisions of "Differential moveout between primary and multiple"

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 6 181 - 220 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 6.10a

A multiple reflection is produced by a horizontal bed at a depth of 1100 m; average velocity to this depth is 2000 m/s. A primary reflection from a depth of 3250 m coincides with the multiple at zero offset. By how much do arrival times differ at points 200, 400, 800, and 1000 m from the source?

### Solution

Raypaths for the 200 and 1000 m offsets are drawn to scale in Figure 6.10a. We treat this as a two-layer problem with a 1100-m layer over a 2150-m layer, the velocity in the second layer being such that the traveltimes in the two layers are equal. Since ${\bar {V_{1}}}=2000{\rm {m/s}}$ , the travel-time $t_{0}$ in the upper layer is $2\times 1100/2000=1.100\ {\rm {s}}$ and $V_{2}=2\times 2150/1.100=3910\ {\rm {m/s}}$ , the average velocity from the surface being ${\bar {V_{2}}}=2\times 3250/2.200=2950\ {\rm {m/s}}$ .

Assuming a straight-line raypath at the 1000-m offset, we get an angle of ${\rm {tan}}^{-1}(500/3250)\approx 9^{\circ }$ , and hence, the raypath bending will be small and we can ignore it.

The arrival time of the deep reflector is $t_{r}=(6500^{2}+x^{2})^{1/2}/2950$ and the multiple’s $t_{m}=(4400^{2}+x^{2})^{1/2}/2000$ . Their arrival times and differential normal moveouts $\Delta (\Delta t_{\hbox{NMO}})$ are listed in Table 6.10a.

## Problem 6.10b

If the shallow bed dips $10^{\circ }$ , how much do the arrival times at 400 and 800 m change? What is the apparent dip of the multiple?

### Solution

We have two cases to consider: offset updip and offset downdip. We use the notation shown in Figure 6.10b to denote the various angles of incidence and refraction. The offsets and path lengths for the shallow multiple are easily found graphically. We assume that the depths given in part (a) are vertical depths at the source. From part (a) we have $V_{2}/V_{1}=1.96$ . We now calculate angles from the following relations:

{\begin{aligned}a_{1}=\xi -\alpha _{1},a_{1}^{'}={\rm {sin}}^{-1}(1.96{\rm {\;sin\;}}a_{1}),\end{aligned}} {\begin{aligned}\alpha _{2}=\xi -a_{1}^{'},b_{1}^{'}=\xi +\alpha _{2}=2\xi -a_{1}^{'},\end{aligned}} {\begin{aligned}\beta _{1}={\rm {sin}}^{-1}[({\rm {\;sin\;}}b_{1}^{'})/1.96].\end{aligned}} Table 6.10a. Arrival times and differential NMO.
$x({\hbox{m}})$ $t_{r}({\hbox{s}})$ $t_{m}({\hbox{s}})$ $\Delta (\Delta t_{\rm {NMO}})$ 0 2.200 2.200 0
200 2.204 2.202 0.001
400 2.208 2.209 0.001
600 2.213 2.220 0.007
800 2.220 2.236 0.016
1000 2.229 2.256 0.027
Table 6.10b. Downdip and updip angles, offsets and traveltimes for primary.
Downdip Updip
$\alpha _{1}$ 0.0 $1.0^{\circ }$ $2.0^{\circ }$ $4.9^{\circ }$ $7.0^{\circ }$ $8.0^{\circ }$ $a_{1}$ $10.0^{\circ }$ $9.0^{\circ }$ $8.0^{\circ }$ $5.1^{\circ }$ $3.0^{\circ }$ $2.0^{\circ }$ ${a'}_{1}$ $19.9^{\circ }$ $17.9^{\circ }$ $15.8^{\circ }$ $10.0^{\circ }$ $5.9^{\circ }$ $3.9^{\circ }$ $a_{2}$ $-9.9^{\circ }$ $-7.9^{\circ }$ $-5.8^{\circ }$ $0.0^{\circ }$ $4.1^{\circ }$ $6.1^{\circ }$ $b_{2}$ $-9.9^{\circ }$ $-7.9^{\circ }$ $-5.8^{\circ }$ 0.0 $4.1^{\circ }$ $6.1^{\circ }$ $b_{1}$ $0.1^{\circ }$ $+2.1^{\circ }$ $+4.2^{\circ }$ $10.0^{\circ }$ $14.1^{\circ }$ $16.1^{\circ }$ ${b'}_{1}$ 0.0 $+1.1^{\circ }$ $+2.1^{\circ }$ $5.1^{\circ }$ $7.1^{\circ }$ $8.1^{\circ }$ $x$ 970 m 810 m 630 m 0.00 390 m 790 m
$t_{m}^{d}$ 1.528 s 1.520 s 1.502 s 1.488 s 1.475 s 1.471 s

Next we plot the raypaths and measure the offsets and path lengths and finally calculate the traveltimes. The calculated angles and measured values of $x$ and $t_{r}$ are listed in Table 6.10b for the downdip and updip cases.

The graphical construction and path length measurements are illustrated in Figure 6.10c. The primary arrival times at the required offsets are found by interpolation, using values of $x$ and $t_{r}^{d}$ in Table 6.10c; the results are shown in the first two rows of Table 6.10c. For the multiple we use the method of images (see problem 4.1) to get offsets and path lengths (see Figure 6.10c). Dividing the path lengths by the velocity gives the traveltimes in Table 6.10c.

To find how much the dip has changed the arrival times, we have inserted the zero-dip values $t_{r}$ and $t_{m}$ in Table 6.10c and entered the changes $\Delta t_{r}=t_{r}^{d}-t_{r}$ , $\Delta t_{m}=t_{m}^{d}-t_{m}$ .

 offset $x$ −800 m −400 0.00 400 800 time of dipping reflection $t_{r}^{d}$ 1.520 s 1.502 1.488 1.475 1.471 time of reflection without dip $t_{r}$ 1.501 s 1.494 1.491 1.494 1.501 $t_{r}^{d}-t_{r}$ $\Delta t_{r}$ 0.019 s 0.007 −0.003 −0.018 −0.024 multiple of dipping reflection $t_{m}^{d}$ 1.549 s 1.488 1.437 1.397 1.366 multiple reflection without dip $t_{m}$ 1.510 s 1.498 1.492 1.498 1.510 $t_{m}^{d}-t_{m}.$ $\Delta t_{m}$ −0.039 s −0.010 0.055 0.101 0.144

Thus the changes in both primaries $\Delta t_{r}$ and multiples $\Delta t_{m}$ are significant. To get the apparent dip of the multiple, we use the data for $x=800\ {\rm {m}}$ in Table 6.10c; the time difference is $(1.549-1.366)=0.183\ {\rm {s}}$ . The apparent dip is given by equation (4.2b), assuming ${\bar {V_{2}}}=2950$ so

{\begin{aligned}{\rm {\;sin\;}}\xi _{a}=(2.95/2)(0.183/0.80)=0.337,\xi _{a}=19.7^{\circ }.\end{aligned}} The apparent dip moveout for the deep horizontal reflector when the shallow horizon dips $10^{\circ }$ is $1.519-1.471=0.048\ {\rm {s}}$ for $2\Delta x=1600\ {\rm {m}}$ ; hence it now has the apparent dip given by

{\begin{aligned}{\rm {\;sin\;}}\xi =(2.95/2)(0.048/0.80)=0.089,\ {\rm {or}}\ \xi =5.1^{\circ }.\end{aligned}} 